AIEEE Questions: Graphs and Applications

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a small bright lamp, one metre from a photocell, knocking electrons out of the metal inside. Now you move the lamp closer, to half a metre. Half the distance, so twice as many electrons, right? Hmm, are you sure about that? Hold on to your answer, because an A I triple E question asked exactly this. Today we solve two such A I triple E questions, from the exam that later became JEE Main. Questions like these come up often, so let's begin.

0:36Einstein's equation as a straight line

So, first, a quick recap of Einstein's equation. K max equals h nu minus phi naught. K max, the capital K with the small word max, is the maximum kinetic energy. Here h is Planck's constant, nu, the Greek letter like a v, is the frequency, and phi naught, the Greek letter phi with a small zero, is the work function. And K max also equals e times V naught, where e is the electron's charge, and V naught, the V with a small zero, is the stopping potential. Okay? Now look at the graph on the left: stopping potential up the side, frequency along the bottom, with one line for metal A in dark blue, and one for metal B in orange. The label under them says parallel lines of slope h over e. Let me ask you this: why h over e, and why is it the same for both metals? It is really simple: divide Einstein's equation by e. Then V naught equals h over e, times nu, minus phi naught over e, so the slope is h over e. Neither h nor e depends on the metal, so the two lines are parallel. Right? Each line cuts the frequency axis where V naught is zero, at the threshold frequency, nu naught, where h nu naught equals phi naught. For metal B it is nu naught prime, and the little dash is read as prime. Produced back along the dashes, the lines cut the V naught axis at minus phi A over e, and minus phi B over e, with A and B written small. Clear? And what happens below nu naught? There is no emission at all, however intense the light, because one photon must pay the whole work function. So the real graph starts only at nu naught. Got it? Now the graph on the right: saturation current against intensity, at a fixed frequency above the threshold. It is a straight line through the origin, so the saturation current is proportional to the intensity; the sign that looks like an open alpha means is proportional to. With me?

2:46Intensity, distance and the photocurrent

Now look at this figure, on the left: photocurrent up the side, collector potential along the bottom. The three blue curves are for one frequency but three intensities, capital I one, I two and I three, with I three the brightest. All three meet the axis at the same point, minus V naught, and the label below says the same stopping potential. But brighter light carries more energy, so why do its electrons not need a bigger stopping potential? Because brighter light brings more photons, not more energetic ones, and each photon still gives an electron the same h nu. So the line reads: same nu, I three greater than I two greater than I one, the same V naught. Only the number of electrons goes up, and with it the saturation current. Okay? On the right, the orange curves have the same intensity, but three frequencies, with nu three the highest. Higher frequency gives faster electrons, so a larger stopping potential. The labels are minus V naught one, two and three, with small subscripts, and minus V naught three sits farthest left. One note: the curves are drawn with one saturation current, which takes the same number of photons each second. At the same light power, a higher frequency means fewer photons, so a smaller saturation current. Right? Often the light comes from a point source, of power P source, the P with the small word source. Its light spreads over a sphere of area four pi r squared, and the cathode, of area capital A at a distance r, catches only its share. So P equals P source times capital A, over four pi r squared. Clear? So the saturation current, small i, is proportional to one over r squared. But the stopping potential is unchanged, because each photon still brings the same h nu. Our second question is built on this inverse square law. Got it?

4:47Example 1: Kmax vs frequency graph, slope and intercepts

Now, Example one, an A I triple E question from two thousand four. According to Einstein's photoelectric equation, what is the plot of the kinetic energy of the emitted photoelectrons against the frequency of the incident radiation? Pause the video and try it first. Done? Okay, let's check. The kinetic energy here means K max, and Einstein's equation gives K max equals h nu minus phi naught. Both h and phi naught are fixed for a given metal, so K max is linear in nu. With me? Now the slope. K max equals e times V naught, so multiply the slope of the stopping potential line, h over e, by e. That gives just h, the same for every metal. Right? So K max grows steadily as nu grows. Does the line then pass through the origin, so that K max is proportional to nu? No, no, careful, that is the trap! K max becomes zero when h nu equals phi naught, that is, at nu naught, and not at zero frequency. So the line cuts the frequency axis at the threshold frequency, nu naught. Clear? And if we produce the line back to nu equal to zero, it cuts the energy axis at minus phi naught, a negative intercept, though the real graph starts at nu naught. Got it? So the answer is a straight line of slope h, cutting the frequency axis at nu naught, with a negative intercept on the energy axis. Linear, but not proportional; remember that. Okay?

6:27Example 2: lamp moved from 1 m to 0.5 m, electron count

And now Example two, from A I triple E two thousand five, and it is our puzzle from the start. The question says: a photocell is illuminated by a small bright source placed one metre away, and the m on screen is metre. When the same source is placed half a metre away, how does the number of electrons emitted by the photocathode change? Pause the video and try it first. Done? Okay, let's check. Half the distance, so twice the electrons, right? Oh no, careful, that is the trap! The small source acts like a point source, and its light spreads over a sphere. So the intensity, capital I, is proportional to one over r squared, not one over r. Clear? So I two over I one equals r one over r two, the whole squared. Here I one and I two, with a small one and two, are the intensities at the distances r one and r two. That is one over one half, the whole squared, which is two squared, so the ratio is four. Right? The light already ejects electrons, so it is above the threshold, and moving the lamp does not change its frequency. So the electrons emitted per second are proportional to the intensity: four times the photons free four times the electrons. With me? So the number of electrons increases by a factor of four. There is our puzzle, solved: half the distance gives four times the electrons, not twice. Clear? And notice: each electron is no faster than before, so K max and the stopping potential stay the same, and only the count changes. Got it?

8:10Quick revision: slope h, same V0, current goes as 1 over r2

So, quick revision. K max equals h nu minus phi naught: a straight line of slope h that cuts the frequency axis at nu naught. Brighter light at the same frequency gives more electrons, but the same stopping potential. And for a point source, the saturation current is proportional to one over r squared, so half the distance gives four times the current. Now go back to the lessons on the graphs and on the photocell in this chapter, and try their practice problems. You can do this. All the best!